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a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=3^7.\frac{1}{3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{16}=2^7:2^3.16=2^4.2^4=2^8\)
c) \(3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=2^5.2^2=2^7\)
d) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^3.\left(3^2\right)^2=\frac{1^3}{3^3}.3^4=1^3.3=3^1\)
a) 158 x 94
= 158 x ( 32 )4
= 158 x 38
= ( 15 x 3 )8 = 458
b) 49 : 527
= 49 : ( 53 ) 9
= 49 : 1259
= \(\left(\frac{4}{125}\right)^9\)
c) 2010 : 220
= 2010 : ( 22 )10
= 2010 : 410 = ( 20 : 4 ) 10 = 510
d) 275 : ( -7 ) 15
= 275 : [ ( - 7 )3 ]5
= 275 : ( - 21 )5
= \(\left(\frac{27}{-21}\right)^5=\left(\frac{9}{-7}\right)^5\)
Cbht
a) Ta có :
\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)
\(\Rightarrow27^{27}>243^{13}\)
\(\Rightarrow-27^{27}< -243^{13}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)
b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)
c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)
\(\Rightarrow4^{50}>8^{30}\)
d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)
\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)
a) \(2.4.16.32.2^4=2.2^2.2^4.2^5.2^4=2^{16}\)
b) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=\left(2^2.2^5\right):\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^7:\frac{1}{2}=2^8\)
c) \(9.3^3.\frac{1}{81}.27=3^2.3^3.\left(\frac{1}{3}\right)^4.3^3=3^4\)
d)\(2^2.4.\frac{32}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)
Bài 1 : bạn cứ đóng ngoặc bài lại rồi cho thêm mũ nào đó vào là xong
bài 2:
a,(2x-3)^2=4
(2x-3)^2=(+-2)^2
=> 2x-3=(+-2)
(bn cứ phân ra 2 trường hợp rồi từ từ làm
a: Sửa đề: 3^2
\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^3}=3^2\)
b: \(=3^{\left(-2\right)\cdot\left(-2\right)}\cdot\dfrac{1}{3^5}\cdot3^3=\dfrac{3^4}{3^2}=3^2\)
c: \(=2^{12}\cdot2^{16}\cdot2^4=2^{32}\)
d: \(=\left[\dfrac{1}{9}\cdot\dfrac{27}{8}\cdot3\right]\cdot\dfrac{128}{81}\)
\(=\dfrac{16}{9}=\left(\dfrac{4}{3}\right)^2\)